$\dfrac{dy}{dt}=2t + 3$ and $y(1)=6$. What is $t$ when $y=0$ ? Choose all answers that apply: Choose all answers that apply: (Choice A) A $t=-3$ (Choice B) B $t=-2$ (Choice C) C $t=1$ (Choice D) D $t=0$ (Choice E) E $t=-1$ (Choice F) F $t=-4$
Answer: The differential equation is separable. What does it look like after we separate the variables? $dy= (2t+3)\,dt$ Let's integrate both sides of the equation. $\int dy= \int(2t+3)\,dt$ What do we get? $y=t^2+3t + C$ What value of $C$ satisfies the initial condition $y(1)=6$ ? Let's substitute $t=1$ and $y=6$ into the equation and solve for $C$. $\begin{aligned} 6&=1^2+3\cdot1 + C\\ \\ 6&=4+C\\ \\ C&=2 \end{aligned}$ Now use this value of $C$ to find $t$ when $y=0$. $0=t^2+3t+2$ $0=(t+2)(t+1)$ Thus, $t=-2$ or $t=-1$.